3.2.11 \(\int \frac {x^4}{\sinh ^{-1}(a x)^{7/2}} \, dx\) [111]

Optimal. Leaf size=285 \[ -\frac {2 x^4 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {16 x^3}{15 a^2 \sinh ^{-1}(a x)^{3/2}}-\frac {4 x^5}{3 \sinh ^{-1}(a x)^{3/2}}-\frac {32 x^2 \sqrt {1+a^2 x^2}}{5 a^3 \sqrt {\sinh ^{-1}(a x)}}-\frac {40 x^4 \sqrt {1+a^2 x^2}}{3 a \sqrt {\sinh ^{-1}(a x)}}-\frac {\sqrt {\pi } \text {Erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{30 a^5}+\frac {9 \sqrt {3 \pi } \text {Erf}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{20 a^5}-\frac {5 \sqrt {5 \pi } \text {Erf}\left (\sqrt {5} \sqrt {\sinh ^{-1}(a x)}\right )}{12 a^5}+\frac {\sqrt {\pi } \text {Erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{30 a^5}-\frac {9 \sqrt {3 \pi } \text {Erfi}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{20 a^5}+\frac {5 \sqrt {5 \pi } \text {Erfi}\left (\sqrt {5} \sqrt {\sinh ^{-1}(a x)}\right )}{12 a^5} \]

[Out]

-16/15*x^3/a^2/arcsinh(a*x)^(3/2)-4/3*x^5/arcsinh(a*x)^(3/2)-1/30*erf(arcsinh(a*x)^(1/2))*Pi^(1/2)/a^5+1/30*er
fi(arcsinh(a*x)^(1/2))*Pi^(1/2)/a^5+9/20*erf(3^(1/2)*arcsinh(a*x)^(1/2))*3^(1/2)*Pi^(1/2)/a^5-9/20*erfi(3^(1/2
)*arcsinh(a*x)^(1/2))*3^(1/2)*Pi^(1/2)/a^5-5/12*erf(5^(1/2)*arcsinh(a*x)^(1/2))*5^(1/2)*Pi^(1/2)/a^5+5/12*erfi
(5^(1/2)*arcsinh(a*x)^(1/2))*5^(1/2)*Pi^(1/2)/a^5-2/5*x^4*(a^2*x^2+1)^(1/2)/a/arcsinh(a*x)^(5/2)-32/5*x^2*(a^2
*x^2+1)^(1/2)/a^3/arcsinh(a*x)^(1/2)-40/3*x^4*(a^2*x^2+1)^(1/2)/a/arcsinh(a*x)^(1/2)

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Rubi [A]
time = 0.41, antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 32, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {5779, 5818, 5778, 3389, 2211, 2235, 2236} \begin {gather*} -\frac {\sqrt {\pi } \text {Erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{30 a^5}+\frac {9 \sqrt {3 \pi } \text {Erf}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{20 a^5}-\frac {5 \sqrt {5 \pi } \text {Erf}\left (\sqrt {5} \sqrt {\sinh ^{-1}(a x)}\right )}{12 a^5}+\frac {\sqrt {\pi } \text {Erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{30 a^5}-\frac {9 \sqrt {3 \pi } \text {Erfi}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{20 a^5}+\frac {5 \sqrt {5 \pi } \text {Erfi}\left (\sqrt {5} \sqrt {\sinh ^{-1}(a x)}\right )}{12 a^5}-\frac {16 x^3}{15 a^2 \sinh ^{-1}(a x)^{3/2}}-\frac {40 x^4 \sqrt {a^2 x^2+1}}{3 a \sqrt {\sinh ^{-1}(a x)}}-\frac {2 x^4 \sqrt {a^2 x^2+1}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {32 x^2 \sqrt {a^2 x^2+1}}{5 a^3 \sqrt {\sinh ^{-1}(a x)}}-\frac {4 x^5}{3 \sinh ^{-1}(a x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/ArcSinh[a*x]^(7/2),x]

[Out]

(-2*x^4*Sqrt[1 + a^2*x^2])/(5*a*ArcSinh[a*x]^(5/2)) - (16*x^3)/(15*a^2*ArcSinh[a*x]^(3/2)) - (4*x^5)/(3*ArcSin
h[a*x]^(3/2)) - (32*x^2*Sqrt[1 + a^2*x^2])/(5*a^3*Sqrt[ArcSinh[a*x]]) - (40*x^4*Sqrt[1 + a^2*x^2])/(3*a*Sqrt[A
rcSinh[a*x]]) - (Sqrt[Pi]*Erf[Sqrt[ArcSinh[a*x]]])/(30*a^5) + (9*Sqrt[3*Pi]*Erf[Sqrt[3]*Sqrt[ArcSinh[a*x]]])/(
20*a^5) - (5*Sqrt[5*Pi]*Erf[Sqrt[5]*Sqrt[ArcSinh[a*x]]])/(12*a^5) + (Sqrt[Pi]*Erfi[Sqrt[ArcSinh[a*x]]])/(30*a^
5) - (9*Sqrt[3*Pi]*Erfi[Sqrt[3]*Sqrt[ArcSinh[a*x]]])/(20*a^5) + (5*Sqrt[5*Pi]*Erfi[Sqrt[5]*Sqrt[ArcSinh[a*x]]]
)/(12*a^5)

Rule 2211

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - c*(
f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2236

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F],
2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 3389

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/E^(I*(e + f*x))
, x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 5778

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x^m*Sqrt[1 + c^2*x^2]*((a + b*ArcSi
nh[c*x])^(n + 1)/(b*c*(n + 1))), x] - Dist[1/(b^2*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[x^(n + 1), Si
nh[-a/b + x/b]^(m - 1)*(m + (m + 1)*Sinh[-a/b + x/b]^2), x], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c}
, x] && IGtQ[m, 0] && GeQ[n, -2] && LtQ[n, -1]

Rule 5779

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x^m*Sqrt[1 + c^2*x^2]*((a + b*ArcSi
nh[c*x])^(n + 1)/(b*c*(n + 1))), x] + (-Dist[c*((m + 1)/(b*(n + 1))), Int[x^(m + 1)*((a + b*ArcSinh[c*x])^(n +
1)/Sqrt[1 + c^2*x^2]), x], x] - Dist[m/(b*c*(n + 1)), Int[x^(m - 1)*((a + b*ArcSinh[c*x])^(n + 1)/Sqrt[1 + c^
2*x^2]), x], x]) /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && LtQ[n, -2]

Rule 5818

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] - Dist[f*(m/
(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]], Int[(f*x)^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1), x], x]
/; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {x^4}{\sinh ^{-1}(a x)^{7/2}} \, dx &=-\frac {2 x^4 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}+\frac {8 \int \frac {x^3}{\sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{5/2}} \, dx}{5 a}+(2 a) \int \frac {x^5}{\sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{5/2}} \, dx\\ &=-\frac {2 x^4 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {16 x^3}{15 a^2 \sinh ^{-1}(a x)^{3/2}}-\frac {4 x^5}{3 \sinh ^{-1}(a x)^{3/2}}+\frac {20}{3} \int \frac {x^4}{\sinh ^{-1}(a x)^{3/2}} \, dx+\frac {16 \int \frac {x^2}{\sinh ^{-1}(a x)^{3/2}} \, dx}{5 a^2}\\ &=-\frac {2 x^4 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {16 x^3}{15 a^2 \sinh ^{-1}(a x)^{3/2}}-\frac {4 x^5}{3 \sinh ^{-1}(a x)^{3/2}}-\frac {32 x^2 \sqrt {1+a^2 x^2}}{5 a^3 \sqrt {\sinh ^{-1}(a x)}}-\frac {40 x^4 \sqrt {1+a^2 x^2}}{3 a \sqrt {\sinh ^{-1}(a x)}}+\frac {32 \text {Subst}\left (\int \left (-\frac {\sinh (x)}{4 \sqrt {x}}+\frac {3 \sinh (3 x)}{4 \sqrt {x}}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{5 a^5}+\frac {40 \text {Subst}\left (\int \left (\frac {\sinh (x)}{8 \sqrt {x}}-\frac {9 \sinh (3 x)}{16 \sqrt {x}}+\frac {5 \sinh (5 x)}{16 \sqrt {x}}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^5}\\ &=-\frac {2 x^4 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {16 x^3}{15 a^2 \sinh ^{-1}(a x)^{3/2}}-\frac {4 x^5}{3 \sinh ^{-1}(a x)^{3/2}}-\frac {32 x^2 \sqrt {1+a^2 x^2}}{5 a^3 \sqrt {\sinh ^{-1}(a x)}}-\frac {40 x^4 \sqrt {1+a^2 x^2}}{3 a \sqrt {\sinh ^{-1}(a x)}}-\frac {8 \text {Subst}\left (\int \frac {\sinh (x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{5 a^5}+\frac {5 \text {Subst}\left (\int \frac {\sinh (x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^5}+\frac {25 \text {Subst}\left (\int \frac {\sinh (5 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{6 a^5}+\frac {24 \text {Subst}\left (\int \frac {\sinh (3 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{5 a^5}-\frac {15 \text {Subst}\left (\int \frac {\sinh (3 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^5}\\ &=-\frac {2 x^4 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {16 x^3}{15 a^2 \sinh ^{-1}(a x)^{3/2}}-\frac {4 x^5}{3 \sinh ^{-1}(a x)^{3/2}}-\frac {32 x^2 \sqrt {1+a^2 x^2}}{5 a^3 \sqrt {\sinh ^{-1}(a x)}}-\frac {40 x^4 \sqrt {1+a^2 x^2}}{3 a \sqrt {\sinh ^{-1}(a x)}}+\frac {4 \text {Subst}\left (\int \frac {e^{-x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{5 a^5}-\frac {4 \text {Subst}\left (\int \frac {e^x}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{5 a^5}-\frac {5 \text {Subst}\left (\int \frac {e^{-x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{6 a^5}+\frac {5 \text {Subst}\left (\int \frac {e^x}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{6 a^5}-\frac {25 \text {Subst}\left (\int \frac {e^{-5 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{12 a^5}+\frac {25 \text {Subst}\left (\int \frac {e^{5 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{12 a^5}-\frac {12 \text {Subst}\left (\int \frac {e^{-3 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{5 a^5}+\frac {12 \text {Subst}\left (\int \frac {e^{3 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{5 a^5}+\frac {15 \text {Subst}\left (\int \frac {e^{-3 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{4 a^5}-\frac {15 \text {Subst}\left (\int \frac {e^{3 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{4 a^5}\\ &=-\frac {2 x^4 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {16 x^3}{15 a^2 \sinh ^{-1}(a x)^{3/2}}-\frac {4 x^5}{3 \sinh ^{-1}(a x)^{3/2}}-\frac {32 x^2 \sqrt {1+a^2 x^2}}{5 a^3 \sqrt {\sinh ^{-1}(a x)}}-\frac {40 x^4 \sqrt {1+a^2 x^2}}{3 a \sqrt {\sinh ^{-1}(a x)}}+\frac {8 \text {Subst}\left (\int e^{-x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{5 a^5}-\frac {8 \text {Subst}\left (\int e^{x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{5 a^5}-\frac {5 \text {Subst}\left (\int e^{-x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{3 a^5}+\frac {5 \text {Subst}\left (\int e^{x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{3 a^5}-\frac {25 \text {Subst}\left (\int e^{-5 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{6 a^5}+\frac {25 \text {Subst}\left (\int e^{5 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{6 a^5}-\frac {24 \text {Subst}\left (\int e^{-3 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{5 a^5}+\frac {24 \text {Subst}\left (\int e^{3 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{5 a^5}+\frac {15 \text {Subst}\left (\int e^{-3 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{2 a^5}-\frac {15 \text {Subst}\left (\int e^{3 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{2 a^5}\\ &=-\frac {2 x^4 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {16 x^3}{15 a^2 \sinh ^{-1}(a x)^{3/2}}-\frac {4 x^5}{3 \sinh ^{-1}(a x)^{3/2}}-\frac {32 x^2 \sqrt {1+a^2 x^2}}{5 a^3 \sqrt {\sinh ^{-1}(a x)}}-\frac {40 x^4 \sqrt {1+a^2 x^2}}{3 a \sqrt {\sinh ^{-1}(a x)}}-\frac {\sqrt {\pi } \text {erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{30 a^5}+\frac {9 \sqrt {3 \pi } \text {erf}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{20 a^5}-\frac {5 \sqrt {5 \pi } \text {erf}\left (\sqrt {5} \sqrt {\sinh ^{-1}(a x)}\right )}{12 a^5}+\frac {\sqrt {\pi } \text {erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{30 a^5}-\frac {9 \sqrt {3 \pi } \text {erfi}\left (\sqrt {3} \sqrt {\sinh ^{-1}(a x)}\right )}{20 a^5}+\frac {5 \sqrt {5 \pi } \text {erfi}\left (\sqrt {5} \sqrt {\sinh ^{-1}(a x)}\right )}{12 a^5}\\ \end {align*}

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Mathematica [A]
time = 0.46, size = 334, normalized size = 1.17 \begin {gather*} \frac {-2 e^{\sinh ^{-1}(a x)} \left (3+2 \sinh ^{-1}(a x)+4 \sinh ^{-1}(a x)^2\right )+9 e^{3 \sinh ^{-1}(a x)} \left (1+2 \sinh ^{-1}(a x)+12 \sinh ^{-1}(a x)^2\right )-e^{5 \sinh ^{-1}(a x)} \left (3+10 \sinh ^{-1}(a x)+100 \sinh ^{-1}(a x)^2\right )+100 \sqrt {5} \left (-\sinh ^{-1}(a x)\right )^{5/2} \Gamma \left (\frac {1}{2},-5 \sinh ^{-1}(a x)\right )-108 \sqrt {3} \left (-\sinh ^{-1}(a x)\right )^{5/2} \Gamma \left (\frac {1}{2},-3 \sinh ^{-1}(a x)\right )+8 \left (-\sinh ^{-1}(a x)\right )^{5/2} \Gamma \left (\frac {1}{2},-\sinh ^{-1}(a x)\right )+e^{-\sinh ^{-1}(a x)} \left (-6+4 \sinh ^{-1}(a x)-8 \sinh ^{-1}(a x)^2+8 e^{\sinh ^{-1}(a x)} \sinh ^{-1}(a x)^{5/2} \Gamma \left (\frac {1}{2},\sinh ^{-1}(a x)\right )\right )+9 e^{-3 \sinh ^{-1}(a x)} \left (1-2 \sinh ^{-1}(a x)+12 \sinh ^{-1}(a x)^2-12 \sqrt {3} e^{3 \sinh ^{-1}(a x)} \sinh ^{-1}(a x)^{5/2} \Gamma \left (\frac {1}{2},3 \sinh ^{-1}(a x)\right )\right )+e^{-5 \sinh ^{-1}(a x)} \left (-3+10 \sinh ^{-1}(a x)-100 \sinh ^{-1}(a x)^2+100 \sqrt {5} e^{5 \sinh ^{-1}(a x)} \sinh ^{-1}(a x)^{5/2} \Gamma \left (\frac {1}{2},5 \sinh ^{-1}(a x)\right )\right )}{240 a^5 \sinh ^{-1}(a x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/ArcSinh[a*x]^(7/2),x]

[Out]

(-2*E^ArcSinh[a*x]*(3 + 2*ArcSinh[a*x] + 4*ArcSinh[a*x]^2) + 9*E^(3*ArcSinh[a*x])*(1 + 2*ArcSinh[a*x] + 12*Arc
Sinh[a*x]^2) - E^(5*ArcSinh[a*x])*(3 + 10*ArcSinh[a*x] + 100*ArcSinh[a*x]^2) + 100*Sqrt[5]*(-ArcSinh[a*x])^(5/
2)*Gamma[1/2, -5*ArcSinh[a*x]] - 108*Sqrt[3]*(-ArcSinh[a*x])^(5/2)*Gamma[1/2, -3*ArcSinh[a*x]] + 8*(-ArcSinh[a
*x])^(5/2)*Gamma[1/2, -ArcSinh[a*x]] + (-6 + 4*ArcSinh[a*x] - 8*ArcSinh[a*x]^2 + 8*E^ArcSinh[a*x]*ArcSinh[a*x]
^(5/2)*Gamma[1/2, ArcSinh[a*x]])/E^ArcSinh[a*x] + (9*(1 - 2*ArcSinh[a*x] + 12*ArcSinh[a*x]^2 - 12*Sqrt[3]*E^(3
*ArcSinh[a*x])*ArcSinh[a*x]^(5/2)*Gamma[1/2, 3*ArcSinh[a*x]]))/E^(3*ArcSinh[a*x]) + (-3 + 10*ArcSinh[a*x] - 10
0*ArcSinh[a*x]^2 + 100*Sqrt[5]*E^(5*ArcSinh[a*x])*ArcSinh[a*x]^(5/2)*Gamma[1/2, 5*ArcSinh[a*x]])/E^(5*ArcSinh[
a*x]))/(240*a^5*ArcSinh[a*x]^(5/2))

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Maple [F]
time = 6.12, size = 0, normalized size = 0.00 \[\int \frac {x^{4}}{\arcsinh \left (a x \right )^{\frac {7}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/arcsinh(a*x)^(7/2),x)

[Out]

int(x^4/arcsinh(a*x)^(7/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arcsinh(a*x)^(7/2),x, algorithm="maxima")

[Out]

integrate(x^4/arcsinh(a*x)^(7/2), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arcsinh(a*x)^(7/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{\operatorname {asinh}^{\frac {7}{2}}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/asinh(a*x)**(7/2),x)

[Out]

Integral(x**4/asinh(a*x)**(7/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arcsinh(a*x)^(7/2),x, algorithm="giac")

[Out]

integrate(x^4/arcsinh(a*x)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4}{{\mathrm {asinh}\left (a\,x\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/asinh(a*x)^(7/2),x)

[Out]

int(x^4/asinh(a*x)^(7/2), x)

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